# Locally path connected and connected imply path connected

Let $$X$$ be a connected locally path connected space. I want to show that it is also path connected.

Following a suggestion in this answer, fix $$a\in X$$ and consider the set

$$U_a = \{x\in X : \text{there is a path from } x \text{ to } a\}$$

and show that it is clopen.

If the set is clopen, then we can decompose $$X$$ as a disjoint union of the open sets $$U_a$$ and $$X\setminus U_a$$. Since $$X$$ is connected and $$U_a \not= \varnothing$$ ($$a\in U_a$$), it follows that $$U_a = X$$.

My question concerns how to show that $$U_a$$ is clopen. I considered writing $$U_a$$ and $$X\setminus U_a$$ as unions of open sets as follows:

$$U_a = \bigcup \{\text{path-connected open neighbourhoods of a}\}$$

$$X\setminus U_a = \bigcup_{x\in X}\bigcup\{\text{path-connected open neighbourhoods of x without a}\}$$

It is clear that any path-connected neighbourhood of $$a$$ is a subset of $$U_a$$, but if there is a path from $$x$$ to $$a$$, does it follow that $$x$$ is in a path connected neighbourhood of a?

• I think it is easier to show that every element in $U_a$ has a neighbourhood that is contained in $U_a$, and the same thing for the complement. – asdq Apr 30 at 17:09

Any $$x \in U_a$$ has a path-connected neighborhood, say $$V$$. Any point in $$V$$ can be connected by a path to $$a$$ by first connecting it to $$x$$, and concatenating with the path to $$a$$. Hence, $$V \subset U_a$$, so $$U_a$$ is open.
If $$x \notin U_a$$, then there is a path-connected neighborhood $$W$$ of $$x$$ such that no point of $$W$$ connects to $$a$$ via a path; otherwise we could join that path to $$x$$, so $$x \in U_a$$. Therefore, $$W \subset X\setminus U_a$$, so $$X\setminus U_a$$ is open.